A: Diodes peaks current can exceed the forward steady sate current by a big factor for a short time without destroying itself. It will destroy itself if the power dissipation is exceeded. The answer to your question is the peak voltage has absolutely nothing to do with a 12 volts power supply or any other voltage power supply it has to do with the diode itself
2x the peak supply voltage!
.6v is the peak inverse voltage drop of a silicon made diode. ANSWER: A diode will have a .6 to .7 volts drop [depending on current] in the FORWARD conduction mode. In the inverse it could be 50 to 1000 volts depends on the diode.
You mean peak inverse voltage.It is the maximum voltage (peak) the diode can be reversed biased (inverse) by without being destroyed.
If you want to find it on your own, apply an increasing reverse voltage to the diode until a large current flows. If you do this, put some sort of resistor in series to prevent overload. This will ruin the diode.Or you can check the data sheet. a good search for the particular diode you are using should help, or you can go to the company who sold you the diode and they should have a data sheet readily available. There should be a graph showing reverse current vs. voltage. There should also be a line in the data sheet that says peak reverse voltage which may be easier for you to follow and understand.
Peak power is the highest rated output of the speaker when loud (or peak) parts hit. Continuous/nominal power is the average output at normal and continuous listening levels. If you have the Peak Power rating, you can figure out the nominal level by multiplying the peak power by the square root of two (0.707). Peak Power * 0.707 = Nominal Power.
2x the peak supply voltage!
A: Peak current
The IN5408 diode is an ordinary silicon diode. It has a 3 amp forward current rating, and a 1000 volt peak reverse voltage rating.
Yes you can. The UF is a significantly faster diode. The peak forward surge current is 50 amps less, but apart from that they have much the same characteristics.
Without knowing the circuit this cannot be determined. Sorry.
A capacitor resists a change in voltage (dv/dt = i/c). An inductor resists a change in current (dl/dt = vl). Together, a capacitor and inductor make a tuned circuit. Usually, in a linear power supply, there is a capacitor in parallel with an inductor in series, and often, in a pi filter, another capacitor in parallel. This reduces the peak to peak voltage at the output. It is also possible to put an inductor in series with the rectifier diode, as as to reduce inrush current. In a switching power supply, things are a little bit different. The primary inductor is a current pump, maintaining constant current flow to the load, controlled by the pulse-width oscillator which switches between on-current from source and off-current from schottky diode. The capacitor in this case filters the output, so as to reduce high frequency harmonics.
A diode will provide saturating current if it reaches its Piv. What it should be well at least 1.5 of the applied reverse voltage. For AC at least 2 times to insure that the peak REVERSE voltage is blocked
Peak reverse voltage of a diode is the maximum reverse bias voltage can be applied to diode which does not cause break down.
A: A power supply when designed it will have some nominal criteria and also some maximum output power available which reflect to the peak load that it can stand without burning up
As close to virtually zero volts as is possible: the current will very small but there will be current. This depends on two things: the material that is used to make the diode, and whether the voltage is applied in a forward or reverse direction to the diode. A typical silicon diode will pass forward current above 0.6 V and pass no reverse current until a much higher voltage is applied (check the rated peak inverse voltage PIV)
peak rating
v peak = v rms times square root (2) Note: A junction voltage of 0.3V is atypical. Normally a silicon diode has a forward voltage between 0.6 volts to 1.4V depending on current. Are you sure about the forward voltage? Perhaps you are talking about germanium or schottky diodes?